Yes I do; 1 line typo .. BAAM!!

Mine was reading the statement again "The total time , including the descend and ascend". I don't know if yours failed from the same cause;

You can divide all the brackets sequence into 4 cases

1. Perfect sequence
2. Sequence that lack X leading '(' but no excess '('
3. Sequence that has Y excess '(' but does not lack any leading '('
4. Sequence that lack both Z opening '(' and ZZ following ')'

For case 4 it's obvious that you can not make a 'pair' out of it because it needs at least 2 more sequence to complete it.

For case 1 it must match with another case 1 sequence (Perfect+Perfect)

For case 2 and 3 we just need to match the sequences that X = Y

How to find the answer I'll left as exercise :p

PS. Just a reminder(I think you know this already). To find the case you cannot just take number of '(' — number of ')'. For example, ')(' is not case 1 , but a case 4 because it lack 1 opening and have 1 excess. '(()' is case 3 with Y=1. '())' is case 2 with X=1.

JEBAITED XD

I'm ready for -100 rating XD . Gratz to all who performs well.

My solution is based on dijkstra's algorithm from vertex 1. So while we're running dijkstra , i just pick the first K node(that is not 1) that we visit by dijkstra , if we were about to exceed K just break the loop. (FYI my dijkstra is implemented in a while loop with a priority_queue as the main data structure , you can see the submission from my profile.)

I used something bfs-liked instead , I have the list of last-minimum block at that length then(For each row there can only be at most 1 element so this list will never exceed n) name A.

Considering this example

bcc
caa
ggg

We know that the first character must be 'b' , second character must be 'c' , BUT WE DO NOT KNOW IF THE 'c' IS FROM (1,2) or (2,1) , so after we append 'c' to answer our A must contain both (1,2) and (2,1) for future use.

For the next length each block in A can generate 2 more blocks. For the sake of simplicity i used std::set name B to store it to get rid of duplicate blocks and auto sort for me.Then the next character of the answer is value of the first block in the set(set sorted for me). So I get all the blocks from B which has the value equal to the minimum block back into A and repeat until we reached the answer's length.

This way I do not need to store all the string length but only 1 letter for each block in list A. So the total memory used for this step never exceeds O(n) I guess...

Sorry for my bad English

Maybe I should study more math before participating future contests as of late so many DIV2 C,D are using maths. So the algorithms I know are not needed here. I wish I am a math guy lol.

Magic does exist!! Wth? how did it passed.......

They both rejected. Pretests are just pretests you know.

GG me lost :( Every time I gets near that purple bar I always go haywire lol. Curse of the Expert is strong with me.

Why so much math and greedy? I don't know about E and F though. I solved E by greedy that i did't even understand why it passed the pretest ?_?

The speed is really impressive , I'm adding report to my blog. That's really FAST AND FURIOUS!! Thanks for your awesome code :)

Auto comment: topic has been updated by PoomrokC (previous revision, new revision, compare).

Auto comment: topic has been updated by PoomrokC (previous revision, new revision, compare).

Saddest round of the year , I always fail the Goodbye Round(3 years straight)

Is there any variable that is used uninitialized? Different compiler on different machines will assign different values to those variables;

So what's the real solution for C please , I'm pretty sure i'll be around 1899 because of this C haha.

mine too :)

What should be the answer?

What's the C hacking case was please?

Don't give up yet!! You can do it!!

Is Codeforces trying to implement "Boxing Day" algorithm from the England Premier League? Xp 4 Codeforces round in 3 days. That's something worth living for...

cost me 3 wa

Hint: DP on tree

Example case 2 is your friend , personally i think this problem is too mathematic.

I think there were some mathematical thing that says what it want after the word pair. It's quite clear to me tho.

BINGO!! Nice reason for downvotes. I tried to find rules about writing blog and no rule that said it should only be about cp. I mean , it's a coding community right? Some people may find it useful,others don't. But okay,no more blogs :) . HOPE YOU GUYS HAPPILY LIVE IN THE NICE CP WORLD AND KEEP DOWNVOTING :)

I already used Cmake. You should try it sometimes , follow the official documentation and all the errors come up all the time. The official tutorial also never mention that we can get it through apt-get, that's why I wrote this guide. You are already so smart , you can just skip this guide :)

Auto comment: topic has been updated by PoomrokC (previous revision, new revision, compare).

My solution for C got one line in wrong ordered which should give wrong answer to many of the tests but somehow passed the pretest then system test got me. :( Now I cannot tell whether I'm extremely unlucky or extremely lucky. Xp

//wrong one which pass the pretest
a[i]=*needer.rbegin();
needer.pop_back();
give[a[i]]--;
//right one that i figured out after the contest
give[a[i]]--;
a[i]=*needer.rbegin();
needer.pop_back();

FEELSBADMAN

Will system test start right away? Or With some delay? Thanks in advance

It takes the lastest submission in case you found bug(ssss), some godlike cases that may not be verified in pretests.

When will the system test start please? So I can decide whether i will sleep before school tomorrow Xp(only 3 hours before I must wake up again) Haha

Something strange.... int this codeforce round.... WHO TO CALL??????

(Just a relaxing comment,don't take it so seriously Xp)

Take a look at my graph and keep fighting bro!! We all have bad days :)

That hash table hit me HARD!! What a great lesson by the problem writers :). Thank you, you guys make me learn something new.

Let's hope this be a nice round for everyone :D

Seems like not much people are participating compare to other rounds,that means harder to climb rating Xp. But I will participate anyway. Hope this contest go well,nice and smooth for the round organizer and participants :)